Question: The polynomial $p(x)=2x^3-x^2-25x-12$ has a known factor of $(x+3)$. Rewrite $p(x)$ as a product of linear factors. $p(x)=$
Answer: We know $(x+3)$ is a factor of $p(x)$. This means that $p(x)=(x+3)\cdot q(x)$ for some polynomial $q(x)$. We can find $q(x)$ using polynomial division, and then we can factor $q(x)$. This way, we will be able to rewrite $p(x)$ as a product of linear factors. Dividing $p(x)$ by $(x+3)$ $\begin{array}{r} 2x^2-\phantom{1}7x-\phantom{1}4 \\ x+3|\overline{2x^3-\phantom{6}x^2-25x-12} \\ \mathllap{-(}\underline{2x^3+6x^2\phantom{-58x-12}\rlap )} \\ -7x^2-25x-12 \\ \mathllap{-(}\underline{-7x^2-21x\phantom{-12}\rlap )} \\ -4x-12 \\ \mathllap{-(}\underline{-4x-12\rlap )} \\ 0 \end{array}$ We find that $q(x)=2x^2-7x-4$. Factoring $q(x)$ We can factor $q(x)$ by grouping: $\begin{aligned} q(x)&=2x^2-7x-4 \\\\ &=2x^2-8x+1x-4 \\\\ &=2x(x-4)+1(x-4) \\\\ &=(2x+1)(x-4) \end{aligned}$ Putting it all together $\begin{aligned} p(x)&=2x^3-x^2-25x-12 \\\\ &=(x+3)(2x^2-7x-4) \\\\ &=(x+3)(2x+1)(x-4) \end{aligned}$